Simplify. Multiply and remove all perfect squares from inside the square roots. Assume $y$ is positive. $6\sqrt{15y^4}\cdot 2\sqrt{20y^2}=$
Answer: Let's start by multiplying the factors within and without the square roots: $\begin{aligned} 6\sqrt{15y^4}\cdot 2\sqrt{20y^2} &=6\cdot 2\cdot\sqrt{15y^4}\cdot\sqrt{20y^2} \\\\ &=12\sqrt{300y^6} \end{aligned}$ Now we remove all perfect squares from inside the square root: $\begin{aligned} 12\sqrt{300y^6}&=12\sqrt{10^2\cdot \left(y^3 \right)^2\cdot3} \\\\ &=12\sqrt{10^2}\cdot\sqrt{\left(y^3 \right)^2}\cdot\sqrt{3} \\\\ &=12\cdot 10\cdot y^3\sqrt{3} \\\\ &=120y^3\sqrt{3} \end{aligned}$ In conclusion, $6\sqrt{15y^4}\cdot 2\sqrt{20y^2}=120y^3\sqrt{3}$